How do you factor $1 - 4b² + a² - 2ab$ ?

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Answer 1 · 2016-04-27T20:32:09

This expression does not factorise.

Given: $1-4b^2+a^2-2ab$

Notice that the expression is a mixture of terms of degree $2$ and degree $0$ .

If it did factor then the factors would be a mixture of terms of degree $1$ and $0$ .

The factors can be scaled so that the coefficient of $a$ is $1$ in both factors, resulting in a factorisation of the form:

$1-4b^2+a^2-2ab$

$=(a+pb+r)(a+qb+s)$

$=a^2+(p+q)ab+pqb^2+(r+s)a+(ps+qr)b+rs$

Equating coefficients, we find:

${(p+q=-2), (pq=-4), (r+s=0), (ps+qr=0), (rs=1) :}$

From the first two equations, we find that $p$ and $q$ are (in some order):

${ (-1+sqrt(5)), (-1-sqrt(5)) :}$

From the third and fifth equation, we find that $r$ and $s$ are (in some order):

${ (i), (-i) :}$

Then we find that:

$ps+qr = +-2sqrt(5)i$

contradicting the fourth equation.

So there is no such factorisation.

How do you factor 1 - 4b² + a² - 2ab1 - 4b² + a² - 2ab?